Let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Prove that $x^2yz+y^2zx+z^2xy\le \frac13$

Question

Let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Prove that $x^2yz+y^2zx+z^2xy\le \dfrac13$

My attempt:
I tried using AM-GM and also weighted AM-GM but both seems to be unyielding. So, please help. Thank you.

Answer 1

Apply AM-GM and Cauchy-Schwarz inequalities:

$LHS =xyz(x+y+z) \leq \dfrac{(x+y+z)^4}{27}\leq \dfrac{(3(x^2+y^2+z^2))^2}{27}=\dfrac{1}{3}$

Answer 2

[\begin{gathered} 1 = {x^2} + {y^2} + {z^2} \geqslant 3\sqrt[3]{{{x^2}{y^2}{z^2}}} \Rightarrow {\left( {xyz} \right)^2} \leqslant {\left( {\frac{1} {3}} \right)^3} = \frac{1} {{27}} \Rightarrow xyz \leqslant \frac{1} {{3\sqrt 3 }} \hfill \\ {x^2}yz + {y^2}xz + {z^2}xy = xyz\left( {x + y + z} \right) \leqslant xyz.\sqrt {{1^2} + {1^2} + {1^2}} .\sqrt {{x^2} + {y^2} + {z^2}} = \sqrt 3 xyz \leqslant \sqrt 3 .\frac{1} {{3\sqrt 3 }} = \frac{1} {3} \hfill \\ \end{gathered} ]